Question

Let $f,g:N\to N$ such that $f(n+1)=f\left(n\right)+f\left(1\right),\forall n\in N$ and $g$ be any arbitrary function. Which of the following statements is $\text{NOT}$ true $?$

• A. If $g$ is onto, then $fog$ is one-one
• B. If $f$ is onto, then $f\left(n\right)=n\forall n\in N$
• C. $f$ is one -one
• D. If $fog$ is one-one, then $g$ is one-one

Answer 1

The correct option is A If $g$ is onto, then $fog$ is one-one

$f(n+1)=f\left(n\right)+f\left(1\right)$
$\Rightarrow f(n+1)-f\left(n\right)=f\left(1\right)\to A.P$ with common difference $=f\left(1\right)$
General term $T_n=f\left(1\right)+(n-1)f\left(1\right)=nf\left(1\right)$
$\Rightarrow f\left(n\right)=nf\left(1\right)$
Clearly $f\left(n\right)$ is one-one.
For $fog$ to be one-one, $g$ must be one-one.
For $f$ to be onto, $f\left(n\right)$ should take all the values of natural numbers.
As $f\left(x\right)$ is increasing, $f\left(1\right)=1$
$\Rightarrow f\left(n\right)=n$
If $g$ is many one, then $fog$ is many one. So “if $g$ is onto, then $fog$ is one-one” is incorrect.